Leetcode(120) Triangle

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Description

Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.

For example, given the following triangle

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[
     [2],
    [3,4],
   [6,5,7],
  [4,1,8,3]
]

The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).

Note:

Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.

解法

一道典型的动态规划题：

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dp[i][j] = min(dp[i-1][j-1], dp[i-1][j]) + A[i][j]

具体代码如下：

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class Solution {
    public int minimumTotal(List<List<Integer>> triangle) {
        int minRes = 0;
        Integer[][] dp = new Integer[triangle.size()][triangle.size()];
        dp[0][0] = triangle.get(0).get(0);
        for (int i = 1; i < triangle.size(); i++) {
            for (int j = 0; j <= i; j++) {
                if (j == 0) {
                    dp[i][j] = dp[i - 1][j] + triangle.get(i).get(j);
                } else if (j == i) {
                    dp[i][j] = dp[i - 1][j - 1] + triangle.get(i).get(j);
                } else {
                    dp[i][j] = (dp[i - 1][j - 1] < dp[i - 1][j] ? dp[i - 1][j - 1] : dp[i - 1][j]) + triangle.get(i).get(j);
                }
            }
        }
        minRes = dp[triangle.size() - 1][0];
        for (int j = 0; j <= triangle.size() - 1; j++) {
            if (dp[triangle.size() - 1][j] < minRes) {
                minRes = dp[triangle.size() - 1][j];
            }
        }
        return minRes;
    }
}

看到网上的大佬可以把二维降成一维的解题：
https://blog.csdn.net/fuxuemingzhu/article/details/82883187